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Consider a miner trapped in a room that contains three doors. Door 1 leads him to freedom after two days of travel; door 2 returns him to his room after a four-day journey; and door 3 returns him to his room after a six-day journey. Suppose at all times he is equally likely to choose any of the three doors, and let \(T\) denote the time it takes the miner to become free. (a) Define a sequence of independent and identically distributed random variables \(X_{1}, X_{2} \ldots\) and a stopping time \(N\) such that $$ T=\sum_{i=1}^{N} X_{i} $$ Note: You may have to imagine that the miner continues to randomly choose doors even after he reaches safety. (b) Use Wald's equation to find \(E[T]\). (c) Compute \(E\left[\sum_{i=1}^{N} X_{i} \mid N=n\right]\) and note that it is not equal to \(E\left[\sum_{i=1}^{n} X_{i}\right]\). (d) Use part (c) for a second derivation of \(E[T]\).

Short Answer

Expert verified
The expected time for the miner to become free is 12 days. This is found by defining a sequence of random variables \(X_i\) and a stopping time \(N\), and using Wald's equation to compute \(E[T]\). We verify our result with conditional expectation, showing that \(E[T] = 3 \cdot 4 = 12\).

Step by step solution

01

Part (a) - Define sequence of random variables and stopping time

Let's define the sequence of independent and identically distributed random variables \(X_i\) representing the time it takes for the miner to reach his room or freedom after each attempt, as follows: $$ X_{i} = \begin{cases} 2 & \text{if door 1 is chosen (freedom); with probability }\frac{1}{3} \\ 4 & \text{if door 2 is chosen; with probability }\frac{1}{3} \\ 6 & \text{if door 3 is chosen; with probability }\frac{1}{3} \end{cases} $$ Now, define the stopping time \(N\) to be the first time the miner chooses door 1 and reaches freedom: $$ N = \min \{ n \ge 1: X_{n} = 2 \}$$
02

Part (b) - Compute expected time using Wald's equation

First, let's find the expected value and variance of a single random variable: $$ E[X] = \frac{1}{3}(2) + \frac{1}{3}(4) + \frac{1}{3}(6) = \frac{12}{3} = 4 $$ Since we have a stopping time and a sequence of i.i.d. random variables, we can use Wald's equation to compute the expected time for the miner to become free: $$ E[T] = E\left[\sum_{i=1}^{N} X_{i}\right] = E[N]E[X] $$ Now, we have an expected value for a geometric distribution: $$ E[N] = \frac{1}{p} $$ Since the probability of the miner selecting door 1 (freedom) is \(\frac{1}{3}\): $$ E[N] = \frac{1}{\frac{1}{3}} = 3 $$ Multiplying \(E[N]\) and \(E[X]\): $$ E[T] = E[N]E[X] = 3 \cdot 4 = 12 $$ So, the expected time for the miner to become free is 12 days.
03

Part (c) - Compute conditional expectation

Let's compute the conditional expectation: $$ E\left[\sum_{i=1}^{N} X_{i} \mid N=n\right] = E\left[\sum_{i=1}^{n} X_{i}\right] \frac{P(N=n)}{P(N=n)} = E\left[\sum_{i=1}^{n} X_{i}\right] $$ Now, note that this conditional expectation is not equal to the sum of individual expectations, as: $$ E\left[\sum_{i=1}^{n} X_{i}\right] = nE[X] = 4n $$
04

Part (d) - Second derivation of expected time

Using the conditional expectation found in part (c), we can derive the expected time in a different way: $$ E[T] = E\left[E\left[\sum_{i=1}^{N} X_{i}\right]\Big|\;N\right] = E\left[nE[X]\right] = E[N]\cdot E[X] $$ We already know \(E[N] = 3\) and \(E[X] = 4\): $$ E[T] = 3 \cdot 4 = 12 $$ This confirms our result from part (b) - the expected time for the miner to become free is 12 days.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stochastic Processes
Stochastic processes are essential mathematical models for understanding systems that evolve over time with inherent randomness. Imagine you're observing a queue in a supermarket, not knowing exactly when the next customer will arrive or how many items they’ll have; that’s where stochastic processes come to play. A stochastic process consists of a sequence of random variables, like the sequence of customers' arrival times, which are unpredictable but can still be analyzed using probabilistic methods.

In the textbook exercise, the miner's journey is an example of a stochastic process. Every door selection is a random event in time, forming a sequence until he achieves freedom. This process continues until a specific condition is met, which in this case is the selection of door 1. Exploring stochastic processes, such as this miner’s journey, allows us to predict future events in uncertain conditions, making them incredibly useful for fields ranging from finance to physics.
Random Variables
Random variables are the building blocks of stochastic processes. A random variable isn’t quite 'variable' in the way you might think; it’s a function that assigns a numerical value to every possible outcome of a certain experiment or random event. It’s 'random' because the outcome depends on chance.

In our miner’s example, the number of days it takes to return to his room or reach freedom is a random variable, labeled as \(X_i\). With every door the miner chooses, a value of 2, 4, or 6 days is assigned based on chance. The concept of random variables helps us to understand and quantify uncertainty, allowing for the computation of probabilities and expected values in complex situations. As a powerful toolkit in probability theory, they enable precise predictions and reliable analyses of random phenomena.
Wald's Equation
Wald's equation is a critical principle in stochastic processes and probability theory, which states that the expected value of the sum of a random number of random variables is the product of the expected number of terms and the expected value of a term. This is assuming that the random variables are independent and identically distributed (i.i.d.), paired with a 'stopping time' – a certain condition defining when to stop.

In the context of our exercise, where the miner selects doors until reaching freedom, Wald's equation applies perfectly. It simplifies the process of finding \(E[T]\), the expected total time to reach freedom, which turns out to be the product of the expected time for a single selection \(E[X]\) and the expected number of selections \(E[N]\). Wald's equation proves invaluable for reducing complex problems into manageable calculations.
Expected Value
The expected value, denoted as \(E[X]\), provides a measure of the center of a distribution of a random variable. Think of it like the 'long-term average' if you repeated an experiment over and over again. It's a fundamental concept in probability that captures the idea of the average outcome one would expect.

When we computed \(E[T]\), the expected total time for the miner to find freedom, we made use of the expected value of a single door selection, which is the average time taken for many attempts. Understanding the expected value is key as it gives us insights into the 'typical' result of a stochastic process or a random variable over time, helping with predictions and strategic decision-making.
Geometric Distribution
The geometric distribution records the number of Bernoulli trials needed for a success to occur in a sequence of independent trials, with constant probability of success. It is discrete, meaning it's applicable to situations where you can count the number of trials until success, like tossing a fair coin until heads appears.

In our miner's adventure, the number of door attempts until finding the exit (door 1) follows a geometric distribution with a success probability \(p = \frac{1}{3}\). Consequently, the expected number of attempts to reach success (freedom) is \(E[N] = \frac{1}{p} = 3\), which agrees with the probability theory. Since the geometric distribution only requires a single success to stop, it perfectly models scenarios where one is interested in the probability of 'how soon' or 'how many tries until' an event occurs.

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Most popular questions from this chapter

Each time a certain machine breaks down it is replaced by a new one of the same type. In the long run, what percentage of time is the machine in use less than one year old if the life distribution of a machine is (a) uniformly distributed over \((0,2) ?\) (b) exponentially distributed with mean \(1 ?\)

Consider a single-server queueing system in which customers arrive in accordance with a renewal process. Each customer brings in a random amount of work, chosen independently according to the distribution \(G .\) The server serves one customer at a time. However, the server processes work at rate \(i\) per unit time whenever there are \(i\) customers in the system. For instance, if a customer with workload 8 enters service when there are three other customers waiting in line, then if no one else arrives that customer will spend 2 units of time in service. If another customer arrives after 1 unit of time, then our customer will spend a total of \(1.8\) units of time in service provided no one else arrives. Let \(W_{i}\) denote the amount of time customer \(i\) spends in the system. Also, define \(E[W]\) by $$ E[W]=\lim _{n \rightarrow \infty}\left(W_{1}+\cdots+W_{n}\right) / n $$ and so \(E[W]\) is the average amount of time a customer spends in the system. Let \(N\) denote the number of customers that arrive in a busy period. (a) Argue that $$ E[W]=E\left[W_{1}+\cdots+W_{N}\right] / E[N] $$ Let \(L_{i}\) denote the amount of work customer \(i\) brings into the system; and so the \(L_{i}, i \geqslant 1\), are independent random variables having distribution \(G .\) (b) Argue that at any time \(t\), the sum of the times spent in the system by all arrivals prior to \(t\) is equal to the total amount of work processed by time \(t\). Hint: Consider the rate at which the server processes work. (c) Argue that $$ \sum_{i=1}^{N} W_{i}=\sum_{i=1}^{N} L_{i} $$ (d) Use Wald's equation (see Exercise 13) to conclude that $$ E[W]=\mu $$ where \(\mu\) is the mean of the distribution \(G\). That is, the average time that customers spend in the system is equal to the average work they bring to the system.

A machine in use is replaced by a new machine either when it fails or when it reaches the age of \(T\) years. If the lifetimes of successive machines are independent with a common distribution \(F\) having density \(f\), show that (a) the long-run rate at which machines are replaced equals $$ \left[\int_{0}^{T} x f(x) d x+T(1-F(T))\right]^{-1} $$ (b) the long-run rate at which machines in use fail equals $$ \frac{F(T)}{\int_{0}^{T} x f(x) d x+T[1-F(T)]} $$

In 1984 the country of Morocco in an attempt to determine the average amount of time that tourists spend in that country on a visit tried two different sampling procedures. In one, they questioned randomly chosen tourists as they were leaving the country; in the other, they questioned randomly chosen guests at hotels. (Each tourist stayed at a hotel.) The average visiting time of the 3000 tourists chosen from hotels was \(17.8\), whereas the average visiting time of the 12,321 tourists questioned at departure was \(9.0 .\) Can you explain this discrepancy? Does it necessarily imply a mistake?

Consider a semi-Markov process in which the amount of time that the process spends in each state before making a transition into a different state is exponentially distributed. What kind of process is this?

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